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December 21, 2014

Homework Help: PKA VALUES

Posted by amy on Friday, March 2, 2007 at 4:03pm.

I understand that the pH at the half equivalence point gives is the pKa value but could someone explain why this point is the pKa value?

HA is a weak acid.
HA ==> H^+ + A^-

Ka = (H^+)(A^-)/(HA)
Solve for (H^+)= Ka*(HA)/(A^-)

If we start with, say, 0.1 M HA and start titrating with 0.1 M NaOH, then at the half way point,
1/2 of the 0.1 will have been neutralized leaving exactly 0.05 mols HA remaining. That will have formed exactly 0.05 mols of the salt, NaA, so (HA)=(A^-) so (H^+) = Ka.
Taking the negative log of both sides, leaves
-log(H^+)= - log Ka
pH = pKa.

Neat, huh?

It can also be shown with the Henderson-Hasslebalch equation.
pH = pKa + log (base/(acid)
At the half way point, (base)=(acid) and log 1 = 0, then pH = pKa.

I want to clean up the post a little, mostly for a couple of items omitted.
First, we must take 1 L of the 0.1 M solution of HA. THEN, when we have neutralized exactly half of it, we will have 0.05 mols HA remaining and it will have formed 0.05 mols of the NaA or the anion, A^-. Now the numbers that go into the equation should be molarities, which is mols/L. So If we had 1000 mL (far too much for practical purposes), we will have 0.05 mols HA left and we will have 1000 mL at the beginning + 500 mL of the NaOH that neutralized it and that is a total of 1500 mL. Therefore, the (HA) is 0.05 mols/1.5 L and the (A^-) is 0.05 mols/1.5 L. So (HA)=(A^-), just as my previous post stated and everything works out. I simply didn't want you to confuse molarity (which I started with) with mols (which I ended up with). I left out all of the other logic in between.

Thank you so much that does help a lot. I understood how you went from molarity to mols but thanks for clarifying

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